Question

Two cirele drawn with centres A and B touch each other externally at C, O is a point on the tangent drawn at C, OD and OE are tangents drawn to the two circles of centres A and I respectively. If ∠COD = 56°, ∠COE = 40°, ∠ACD = x° and ∠BCE = y°. Let us Prove that OD = OC = OE and y - x = 8.​

Answer 1

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∠ COD = 56°

∠ COE = 40°

∠ DAC and ∠ DOC are supplementary to each other since AD and AC are perpendicular to OD and OC respectively

⇒ ∠ DAC = (180°-56°) = 124°

Since DA = AC so Δ DAC is isosceles

∠ ACD = 28°

∠ COE and ∠ CBE are supplementary to each other since BC and BE are perpendicular to OC and OE respectively

⇒ ∠ CBE = (180°-40°) = 140°

Since BC = BE so Δ CBE is isosceles

∠ BCE = 20°

y-x = ∠ ACD-∠ BCE = 8°

∠ CDO = ∠ OCD = 62° (∠ ADO and ∠ ACO is equal to 90°)

Hence OD = OC

∠ OEC = ∠ OCE = 70°(∠ OCB and ∠ OEB is equal to 90°)

Hence OE = OC

So combining the above two we can say

OD = OC = OE